Differentiating Rational Powers Revision Notes for VCE SSCE Mathematical Methods

Differentiating Rational Powers

Introduction to rational powers

When working with differentiation, we need to handle not just integer powers like $x^2$ or $x^{-3}$ , but also rational (fractional) powers such as $x^{\frac{1}{2}}$ or $x^{\frac{2}{7}}$ . These are also written using root notation, for example $\sqrt{x} = x^{\frac{1}{2}}$ or $\sqrt[3]{x} = x^{\frac{1}{3}}$ .

To develop our understanding of how to differentiate these functions, we'll first explore the process using first principles, then learn a more efficient method using the chain rule.

Differentiating by first principles

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When we differentiate from first principles, we use the fundamental definition of the derivative:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Let's see how this works for two important fractional powers.

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Worked Example: Differentiate each of the following by first principles:

a) $f(x) = x^{\frac{1}{2}}$ , $x > 0$

b) $g(x) = x^{\frac{1}{3}}$ , $x \neq 0$

Solution

a) For $f(x) = x^{\frac{1}{2}} = \sqrt{x}$ , we start with the difference quotient:

\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h}

To simplify this expression, we multiply by the conjugate:

= \frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}

Using the difference of squares pattern $(a-b)(a+b) = a^2 - b^2$ :

= \frac{x+h-x}{h(\sqrt{x+h} + \sqrt{x})}

= \frac{h}{h(\sqrt{x+h} + \sqrt{x})}

= \frac{1}{\sqrt{x+h} + \sqrt{x}}

Taking the limit as $h \to 0$ :

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}

b) For $g(x) = x^{\frac{1}{3}}$ , we need to use a different algebraic identity. Recall that:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Notice that if we cube both sides of $a^{\frac{1}{3}}$ , we get $a$ . Similarly for $b$ . This means we can write:

a - b = (a^{\frac{1}{3}} - b^{\frac{1}{3}})(a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}})

Rearranging this gives:

a^{\frac{1}{3}} - b^{\frac{1}{3}} = \frac{a-b}{a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}}

Now we can work with the difference quotient:

\frac{g(x+h) - g(x)}{h} = \frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h}

= \frac{x+h-x}{h[(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}]}

= \frac{1}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}

Taking the limit:

g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{1}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}} = \frac{1}{3x^{\frac{2}{3}}}

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A similar identity exists for any positive integer $n$ :

a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1})

We could use this to find the derivative of $x^{\frac{1}{n}}$ from first principles, but the chain rule method we'll see next is more efficient.

Using the chain rule

The reciprocal derivative relationship

The chain rule provides us with a powerful relationship. If $y$ is a one-to-one function of $x$ , then:

\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du}

Setting $y = u$ , we get:

1 = \frac{dy}{dx} \cdot \frac{dx}{dy}

This can be rearranged to give the reciprocal relationship:

\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \quad \text{for } \frac{dx}{dy} \neq 0

This reciprocal relationship is the key to efficiently finding derivatives of fractional powers.

Deriving the formula for $y = x^{\frac{1}{n}}$

Let $y = x^{\frac{1}{n}}$ , where $n \in \mathbb{Z} \setminus \{0\}$ (that is, $n$ is a non-zero integer) and $x > 0$ .

Raising both sides to the power of $n$ :

y^n = x

Now differentiate both sides with respect to $y$ :

\frac{dx}{dy} = ny^{n-1}

Using the reciprocal relationship:

\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{ny^{n-1}}

Substituting $y = x^{\frac{1}{n}}$ back:

\frac{dy}{dx} = \frac{1}{n(x^{\frac{1}{n}})^{n-1}} = \frac{1}{n}x^{\frac{1}{n}-1}

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Formula for fractional powers with unit numerator:

For $y = x^{\frac{1}{n}}$ , we have:

\frac{dy}{dx} = \frac{1}{n}x^{\frac{1}{n}-1}

where $n \in \mathbb{Z} \setminus \{0\}$ and $x > 0$ .

Extending to general rational powers

We can now extend this result to any rational power of the form $x^{\frac{p}{q}}$ , where $p, q \in \mathbb{Z} \setminus \{0\}$ .

Let $y = x^{\frac{p}{q}}$ .

We can write this as:

y = (x^{\frac{1}{q}})^p

Let $u = x^{\frac{1}{q}}$ . Then $y = u^p$ .

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= pu^{p-1} \cdot \frac{1}{q}x^{\frac{1}{q}-1}

Substituting $u = x^{\frac{1}{q}}$ :

= p(x^{\frac{1}{q}})^{p-1} \cdot \frac{1}{q}x^{\frac{1}{q}-1}

= \frac{p}{q}x^{\frac{p}{q} - \frac{1}{q}} \cdot x^{\frac{1}{q}-1}

Combining the powers of $x$ :

= \frac{p}{q}x^{\frac{p}{q}-1}

This shows that the power rule extends to rational powers.

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General power rule for any real power:

For $f(x) = x^a$ , we have:

f'(x) = ax^{a-1}

where $a \in \mathbb{R} \setminus \{0\}$ and $x > 0$ .

In fact, this formula works for any non-zero real number $a$ , not just rational numbers.

Domain and graphs of fractional powers

Why $x > 0$ ?

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The restriction $x > 0$ is important because expressions like $(-3)^{\frac{1}{2}}$ are not defined in the real numbers (since we cannot take the square root of a negative number). However, $(-2)^{\frac{1}{3}}$ is defined because we can take the cube root of a negative number.

To avoid complications, when differentiating fractional powers, we generally restrict the domain to positive real numbers.

Graphs of fractional power functions

The following graph shows three fractional power functions with their domain restricted to $\mathbb{R}^+$ (positive real numbers):

Key observations from the graphs:

All three functions pass through the origin $(0, 0)$
All three functions are increasing
As the denominator of the fractional power increases, the function grows more slowly
$y = x^{\frac{1}{2}}$ grows fastest, then $y = x^{\frac{1}{3}}$ , then $y = x^{\frac{1}{4}}$

Special case: vertical tangent at the origin

The second part of the diagram shows a zoomed-in view of $y = x^{\frac{1}{3}}$ near the origin (range $-0.08 \leq x \leq 0.08$ ).

From this enlarged view, we can see that the tangent to $y = x^{\frac{1}{3}}$ at the origin is vertical (it lies along the $y$ -axis).

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Why is there a vertical tangent?

This makes sense mathematically. The derivative at $x = 0$ would be:

f'(0) = \frac{1}{3}(0)^{\frac{1}{3}-1} = \frac{1}{3}(0)^{-\frac{2}{3}} = \frac{1}{3 \cdot 0^{\frac{2}{3}}}

which is undefined because we're dividing by zero. When the derivative is undefined in this way, it indicates a vertical tangent.

Applying the differentiation rules

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Worked Example: Find the derivative of each of the following with respect to $x$ :

a) $2x^{-\frac{1}{5}} + 3x^{\frac{2}{7}}$

b) $\sqrt[3]{x^2 + 2x}$

Solution

a) We differentiate each term using the power rule:

\frac{d}{dx}(2x^{-\frac{1}{5}} + 3x^{\frac{2}{7}})

For the first term:

\frac{d}{dx}(2x^{-\frac{1}{5}}) = 2 \times (-\frac{1}{5})x^{-\frac{1}{5}-1} = 2 \times (-\frac{1}{5})x^{-\frac{6}{5}} = -\frac{2}{5}x^{-\frac{6}{5}}

For the second term:

\frac{d}{dx}(3x^{\frac{2}{7}}) = 3 \times \frac{2}{7}x^{\frac{2}{7}-1} = 3 \times \frac{2}{7}x^{-\frac{5}{7}} = \frac{6}{7}x^{-\frac{5}{7}}

Therefore:

\frac{d}{dx}(2x^{-\frac{1}{5}} + 3x^{\frac{2}{7}}) = -\frac{2}{5}x^{-\frac{6}{5}} + \frac{6}{7}x^{-\frac{5}{7}}

b) First, rewrite using fractional power notation:

\sqrt[3]{x^2 + 2x} = (x^2 + 2x)^{\frac{1}{3}}

Now we use the chain rule. Let $u = x^2 + 2x$ , so $y = u^{\frac{1}{3}}$ .

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

We have:

\frac{dy}{du} = \frac{1}{3}u^{\frac{1}{3}-1} = \frac{1}{3}u^{-\frac{2}{3}}

\frac{du}{dx} = 2x + 2

Therefore:

\frac{dy}{dx} = \frac{1}{3}u^{-\frac{2}{3}} \times (2x + 2)

Substituting $u = x^2 + 2x$ back:

= \frac{1}{3}(x^2 + 2x)^{-\frac{2}{3}}(2x + 2)

This can be written as:

= \frac{2x + 2}{3\sqrt[3]{(x^2 + 2x)^2}}

= \frac{2x + 2}{3(x^2 + 2x)^{\frac{2}{3}}}

Remember!

bookmarkSummary

Key Points to Remember:

Power rule for rational powers: For any non-zero rational number $r = \frac{p}{q}$ , if $f(x) = x^r$ , then $f'(x) = rx^{r-1}$
Domain restriction: When working with fractional powers, we generally require $x > 0$ to ensure all expressions are defined in the real numbers
Chain rule application: When differentiating composite functions involving rational powers, use the chain rule: differentiate the outer function, then multiply by the derivative of the inner function
Vertical tangents: If the derivative becomes undefined at a point (such as at the origin for $y = x^{\frac{1}{3}}$ ), this may indicate a vertical tangent line
Converting between notations: Remember that $\sqrt[n]{x} = x^{\frac{1}{n}}$ and $\sqrt[n]{x^m} = x^{\frac{m}{n}}$ , which helps when applying the differentiation rules

Differentiating Rational Powers (VCE SSCE Mathematical Methods): Revision Notes